Ionic EquilibriumMedium
Question
The silver ion concentration in a 0.2 M solution of Ag(NH3)2NO3 is (Kdiss = 6.8 × 10−8, 1.53 = 3.4)
Options
A.0.2 M
B.1.5 × 10−3 M
C.1.16 × 10−4 M
D.6.8 × 10−8 M
Solution
$Ag\left( NH_{3} \right)_{2}^{+} \rightleftharpoons Ag^{+} + 2NH_{3}$;Keq = 6.8 × 10–8
0.2 M 0 0
Eqn 0.2 – x = 0.2 M xM 2xM
Now, $6.8 \times 10^{- 8} = \frac{x \times (2x)^{2}}{0.2}$
$\Rightarrow x = 1.5 \times 10^{- 3}\text{ M = }\left\lbrack Ag^{+} \right\rbrack$
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